Discrete random variables

5.4 Discrete random variables

Probability distributions

A discrete random variable takes separate values, each with a probability. A probability distribution lists them, and the probabilities add to 1.

$k + 2k + 3k + 4k = 1$, so $k = 0.1$.

To draw up a table from a situation, list every possible value of $X$ first, then work out each probability. Check that they total 1.

Expectation and variance

$\mathrm{E}(X) = 0.2 + 1.0 + 0.9 = 2.1$

$\sum x^2 p = 0.2 + 2.0 + 2.7 = 4.9$, so $\mathrm{Var}(X) = 4.9 - 2.1^2 = 0.49$.

The binomial distribution

$X \sim \mathrm{B}(n, p)$ counts the number of successes in $n$ trials. It is suitable when:

• There is a fixed number of trials, $n$.
• Each trial has two outcomes, success and failure.
• The trials are independent.
• The probability $p$ of success is constant.

$\mathrm{P}(X = 2) = \binom{10}{2}(0.3)^2(0.7)^8 = 45 \times 0.09 \times 0.05765 = 0.233$

$\mathrm{P}(X \geq 1) = 1 - \mathrm{P}(X = 0) = 1 - 0.7^{10} = 0.972$

Inequalities

Translate the words carefully:

Finding the number of trials

$1 - 0.7^n > 0.95 \quad\Rightarrow\quad 0.7^n < 0.05 \quad\Rightarrow\quad n > \frac{\ln 0.05}{\ln 0.7} = 8.4$

So $n = 9$. The inequality reverses because $\ln 0.7$ is negative.

The geometric distribution

$X \sim \mathrm{Geo}(p)$ counts the number of trials up to and including the first success. There is no upper limit. The trials are independent, with constant probability of success, as for the binomial, but with no fixed number of them.

Getting the first success on trial $r$ needs $r - 1$ failures, then one success.

These two results come from the same idea:

$\mathrm{P}(X > r) = (1 - p)^r \qquad \mathrm{P}(X \leq r) = 1 - (1 - p)^r$

$\mathrm{P}(X > r)$ is the probability that the first $r$ trials are all failures.

$\mathrm{P}(X = 3) = \left(\frac{5}{6}\right)^2\frac{1}{6} = \frac{25}{216} = 0.116$

$\mathrm{P}(X > 4) = \left(\frac{5}{6}\right)^4 = 0.482 \qquad \mathrm{E}(X) = 6$

Which distribution?

When a question combines distributions

A common type of question has two levels: first, the probability for one group, then a binomial on the number of groups.

For one bag: $\mathrm{P}(\text{good}) = 10(0.9)^9(0.1) + (0.9)^{10} = 0.3874 + 0.3487 = 0.7361$.

For the bags, with $\mathrm{B}(3, 0.7361)$:

$\binom{3}{2}(0.7361)^2(0.2639) = 0.429$